\(\int \frac {x^4}{(b x^2+c x^4)^{3/2}} \, dx\) [282]

Optimal result

Integrand size = 19, antiderivative size = 21 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x}{c \sqrt {b x^2+c x^4}} \]

[Out]

-x/c/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {1602} \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x}{c \sqrt {b x^2+c x^4}} \]

[In]

Int[x^4/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x/(c*Sqrt[b*x^2 + c*x^4]))

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {x}{c \sqrt {b x^2+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x}{c \sqrt {x^2 \left (b+c x^2\right )}} \]

[In]

Integrate[x^4/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x/(c*Sqrt[x^2*(b + c*x^2)]))

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38

method	result	size
gosper	\(-\frac {\left (c \,x^{2}+b \right ) x^{3}}{c \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\)	\(29\)
default	\(-\frac {\left (c \,x^{2}+b \right ) x^{3}}{c \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}\)	\(29\)
trager	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}}}{\left (c \,x^{2}+b \right ) c x}\)	\(31\)

[In]

int(x^4/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-(c*x^2+b)/c*x^3/(c*x^4+b*x^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\sqrt {c x^{4} + b x^{2}}}{c^{2} x^{3} + b c x} \]

[In]

integrate(x^4/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

-sqrt(c*x^4 + b*x^2)/(c^2*x^3 + b*c*x)

Sympy [F]

\[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{4}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x**4/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(x**4/(x**2*(b + c*x**2))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {1}{\sqrt {c x^{2} + b} c} \]

[In]

integrate(x^4/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

-1/(sqrt(c*x^2 + b)*c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {\mathrm {sgn}\left (x\right )}{\sqrt {b} c} - \frac {1}{\sqrt {c x^{2} + b} c \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^4/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

sgn(x)/(sqrt(b)*c) - 1/(sqrt(c*x^2 + b)*c*sgn(x))

Mupad [B] (verification not implemented)

Time = 13.74 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {x^4}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {\sqrt {c\,x^4+b\,x^2}}{c\,x\,\left (c\,x^2+b\right )} \]

[In]

int(x^4/(b*x^2 + c*x^4)^(3/2),x)

[Out]

-(b*x^2 + c*x^4)^(1/2)/(c*x*(b + c*x^2))